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Topic: The Most Counterintuitive Probability Paradox? 


This is crazy! (you must watch the video for the full analysis) https://www.youtube.com/watch?v=bDZieLmya_I Scenario # 1: A parent tells you that they have 2 children, and one of them is a girl. What is the probability that the other child is also a girl? Scenario # 2: A parent tells you that they have 2 children, and one of them is a girl, [b][I]and her name is Julie[/I][/b]. Now what is the probability that the other child is also a girl? Apparently, just knowing the girl's name changes the probability of there being 2 girls!! Ron 


I haven't looked at the video, but I presume that it depends on the fact that if they have a second girl, her name isn't Julie. You're correct: this is an interesting situation. One that many, many intelligent people (including engineers, physicists, and mathematicians) get wrong. 


I too find this interesting, and would also have gotten it wrong. I do take exception to one thing. He makes a big point about a boy and girl vs a girl and a boy. I don't think that's relevant to the issue, obviously, he does think it is relevant. 


You asked for itit's turtles all the way down. This has been discussed before by some really amazing folks here including Tomas B and Lobowolf XXX. There are many subthreads referred to as well. Screw your head on tight because it's about to be spinning: https://www.themagiccafe.com/forums/viewtopic.php?topic=321702&forum=101 


[quote]On Apr 8, 2019, Cliffg37 wrote: I too find this interesting, and would also have gotten it wrong. I do take exception to one thing. He makes a big point about a boy and girl vs a girl and a boy. I don't think that's relevant to the issue, obviously, he does think it is relevant.[/quote] Think older boy, younger girl vs. older girl, younger boy. 


[quote]On Apr 8, 2019, landmark wrote: You asked for itit's turtles all the way down. This has been discussed before by some really amazing folks here including Tomas B and Lobowolf XXX. There are many subthreads referred to as well. Screw your head on tight because it's about to be spinning: https://www.themagiccafe.com/forums/viewtopic.php?topic=321702&forum=101 [/quote] Thanks landmark. Wow, this thread goes way back. The video I posted is helpful in the visualization, I think. Anyway, I'm still trying to wrap my brain around it. :) Ron 


At 0:24 he mentions that he was trying "to intuitively understand" what the author of the book had said. Apart from the split infinitive, one cannot "intuitively understand" anything. Intuition means arriving at the correct conclusion [b][i]without[/i][/b] understanding why. I post a lot on a forum for the Chartered Financial Analyst (CFA) exams and have found over the last few years many, many candidates there asking for help "to understand the intuition" behind various various ideas, formulae, and what have you. It's infuriating, and it's caused primarily by prep providers who frequently refer to "the intuition behind" some idea or another when they mean the explanation behind it. I, for one, would much rather have understanding than intuition. In the book [b]Please Understand Me[/b], about the MyersBriggs personality types, the [i]Portrait of an INTP[/i] includes this sentence: [i]INTPs deal with their environment primarily through intuition, and their strongest quality, the thinking function, remains relatively hidden except in close associations.[/i] I disagree. I believe that INTPs deal with their environment primarily through [i]understanding[/i], but that their assessment of new situations is so quick that it appears to be intuition. I'm reminded of Holmes' first encounter with Watson, in which Holmes said, "You've been in Afghanistan, I perceive." It appears intuitive, but later in the book, [b]A Study in Scarlet[/b], Holmes outlines the steps in his thinking process that led to that conclusion. Understanding masquerading as intuition. 


Years ago I attended a lecture by Paul Erdős in which he posed a similar problem: You're playing bridge. [For those of you who are unfamiliar with bridge, first, shame on you, and, second, you deal four hands of 13 cards each, face down, one to yourself, one to your partner who sits opposite you, and one each to your two opponents, one of whom sits to your left and the other of whom sits to your right.] Before you look at your cards, your righthand opponent (RHO) picks up his cards and announces, truthfully, "I have an ace." [Aces are high in bridge.] There is a probability that he has a second ace, and given enough skill, time, and motivation, you could calculate that probability. The next time you deal [in bridge the deal rotates clockwise around the table], before you look at your cards, RHO picks up his cards and announces, truthfully, "I have the ace of spades." Again, there is a probability that he has a second ace, and given enough skill, time, and motivation, you could calculate that probability. The question is: how do those two probabilities compare? Is the first greater than the second, or is the second greater than the first, or are they, in fact, equal? To simplify the understanding, Dr. Erdős proposed an easier problem. Instead of having 52 cards, you have only three: the ace of spades (SA), the ace of hearts (HA), and the king of diamonds (DK). And instead of dealing 13 cards to RHO, you deal only two. Suppose he looks at them and says, truthfully, "I have an ace." What, exactly, has he told you? The answer is . . . nothing! You already knew that he had an ace, because there's only one card that isn't an ace (DK), and he has two cards. So the probability of him having a second ace is ⅓: there are three hands he could hold (SA & HA, SA & DK, HA & DK), and only one of those has two aces. Suppose, instead, that he says, truthfully, "I have the ace of spades." Now he's told you something: he doesn't have the hand comprising exactly the HA and DK. Now there are only two possible hands he could hold (SA & HA, SA & DK), only one of which has a second ace, so the probability in this case is ½. Going back to the original problem, as I opined earlier, it does depend on the assumption that if there are two daughters they're not both named Julie. 


Martin Gardner published one of the earliest variants of the paradox in Scientific American. https://en.wikipedia.org/wiki/Boy_or_Girl_paradox 


Hi All, There is no paradox here, only some fast talking con. Consider this: I've flipped two coins and tell you one of them has turned up "heads." What is the probability the other is a "head?" If you think it's 33%, "Go directly to Jail. Do not pass Go. Do not collect $200." The problem with the "two daughters paradox" is that the explanation of why the answer is 33%, is flawed. In that explanation, you are asked to believe that bb, bg, gb, and gg are four different possible outcomes. Given 100 pairs of siblings and assuming equally likely outcomes, there would be 25 bb, 25 bg, 25 gb, and 25 gg. Eliminate the 25 bb pairs, and you are left with 25 gg pairs divided by 75 pairs with at least one female... 33%. However, the "birth order" of the genders is not important in this probability problem and, therefore, bg and gb should not be considered as two different possible outcomes. All that talk about which was born first or second is just fast talking con to make your brain twist in your head. There are, in reality, only three possible outcomes: 1) "both male," 2) "one of each," and 3) "both female." Give 90 pairs of siblings (just to keep the math simple) and assuming equally likely outcomes, there would be 30 "both male," 30 "one of each," and 30 "both female." Eliminate the 30 "both male" pairs and you are left with 30 "both female" pairs divided by 60 pairs with at least one female... 50%. Regards, Steve 


As complicated as the Original video makes this, I think Steve hit the nail on the head. As I said before I did not take birth order into account as I thought it was not relevant to the question at hand. 


[quote]There are, in reality, only three possible outcomes: 1) "both male," 2) "one of each," and 3) "both female." Give 90 pairs of siblings (just to keep the math simple) and assuming equally likely outcomes, there would be 30 "both male," 30 "one of each," and 30 "both female." Eliminate the 30 "both male" pairs and you are left with 30 "both female" pairs divided by 60 pairs with at least one female... 50%. [/quote] This, I believe, is not correct. It may be true that there are only three outcomes, but they are not equally likely. Go back to the flipping two coins 100 times. Are you saying that two tails will show up as often as one of each? If so, that's not correct. In fact, one of each is twice as likely. 


Hi Landmark, I interpret the question, as given, that the parent has two children at the time the statement was made by the parent: two boys, a boy and a girl, or two girls. The possible outcomes are fixed. Calculating the probability of what children he might have before he has any children is irrelevant. That is, the question does not ask about a future event (i.e., I will have two children and one will be female; what is the probability the other will be female?). In terms of "pennies," there are two pennies on the table in front of me as I type. One is a "tail." What is the probability the other is a "tail?" Regards, Steve 


[quote]On Apr 11, 2019, yachanin wrote: There are, in reality, only three possible outcomes: 1) "both male," 2) "one of each," and 3) "both female." Give 90 pairs of siblings (just to keep the math simple) and assuming equally likely outcomes, there would be 30 "both male," 30 "one of each," and 30 "both female."[/quote] You are absolutely correct: if you assume that they're equal then they're equal. The problem lies in your assumption that they're equally likely; they're not. One of each is twice as likely as both male, and twice as likely as both female. It's that simple. 


[quote]On Apr 12, 2019, yachanin wrote: In terms of "pennies," there are two pennies on the table in front of me as I type. One is a "tail." What is the probability the other is a "tail?"[/quote] Do you know which one is a tail? 


Hi S2000magician, Perhaps trying to put my explanation in the same format as that used by the gentleman in the video was a poor choice and just added some unnecessary confusion. I should have just gone with my explanation in the post above yours in the first place :) Regards, Steve 


Yes, I know which is the tail. 


[quote]On Apr 12, 2019, yachanin wrote: Hi S2000magician, Perhaps trying to put my explanation in the same format as that used by the gentleman in the video was a poor choice and just added some unnecessary confusion. I should have just gone with my explanation in the post above yours in the first place :) Regards, Steve [/quote] Howdy, Steve. The problem with your explanation isn't the format, it's the content. I'm sorry, but you're simply wrong. Try this: instead of two pennies, use a penny and a nickel. Assuming that they're fair coins, what are the (all equally likely) head/tail combinations? 


[quote]On Apr 12, 2019, yachanin wrote: Yes, I know which is the tail.[/quote] That's the problem. 


It seems the problem is one of interpreting what is being asked of the problem solver. I see the question simply as "I have a second child. Is it a male or female." Regards, Steve 


[quote]On Apr 12, 2019, yachanin wrote: It seems the problem is one of interpreting what is being asked of the problem solver. I see the question simply as "I have a second child. Is it a male or female." Regards, Steve[/quote] Absolutely true. If you take a look at the wikipedia link that tommy posted above, you'll see that there's a fair amount of discussion about the proper formulation of the question. As I'm very familiar with the problem, I know that your interpretation is not what the problem solver intended. I'm not saying that your interpretation is unreasonable by any means; only that that's not what was intended. 


Hi $2000magician, It’s not the first time I’ve interpreted things differently than others :) I’ve enjoyed the mental exercise... thanks :) Regards, Steve 


[quote]On Apr 12, 2019, yachanin wrote: Hi $2000magician, It’s not the first time I’ve interpreted things differently than others :)[/quote] Such as my username. It's [b][i]S[/i][/b]2000 (as in Honda's incredible roadster), not [i]$[/i]2000. ;) 


LOL! Sorry about that (and I got it right the first time, but thought I had made a mistake). Maybe it's time for new glasses :) Regards, Steve 


[quote]On Apr 12, 2019, yachanin wrote: Yes, I know which is the tail.[/quote] Then you know whether it's the older penny (the wheat penny) or the younger penny (the Lincoln Memorial penny). See: you're using older and younger even with your pennies. 


There probably aren't many on this forum that remember wheat pennies :) Regards, Steve 


Never mind pennies, I remember farthings. 


[quote]On Apr 15, 2019, tommy wrote: Never mind pennies, I remember farthings.[/quote] I remember pennyfarthings. I own one. 


I remember wampum and trading five Eddie Kranepools for one Micky Mantle. 


Is the small wheel a 1/4 of the big wheel? 


[quote]On Apr 15, 2019, tommy wrote: Is the small wheel a 1/4 of the big wheel?[/quote] Even smaller. I'll measure them and get back to you with a concrete answer. 


[quote]On Apr 15, 2019, tommy wrote: Is the small wheel a 1/4 of the big wheel?[/quote] About ⅓ the diameter, so 1/9 the area (when viewed sideways). 


A farthing was a ¼ of a penny in value  from "fourthing". The predecimalisation British system of coinage was introduced by King Henry II. It was based on the troy system of weighing precious metals. The penny was literally one pennyweight of silver. A pound sterling thus weighed 240 pennyweights, or a pound of sterling silver. 


From what I've looked at on Wikipedia, the most recent farthing had a diameter of 20mm and the quarter farthing 13.5 mm. So the area would be about 45% of the larger. The old farthings were as large as 25mm in diameter, which works out to about 29%. 


There's nothing like a probability paradox to get me out of a manyyearslong spell away from the forum... Here's a Jedi Grandmaster level probability puzzle: Suppose I have an everywherenonzero probability distribution on the real number line. If you don't know what that means, in layman's terms it means that I have some probabilistic process or method to generate a random number. It can potentially generate ANY number. Moreover, if you pick any range of numbers, whether the range is large or small, there's a nonzero probability that you can generate a number inside that range. (It may not be obvious, but such things exist.) But importantly, you don't know anything else about the distribution. You don't know its mean (or if it even has a mean), you don't know if it has any symmetries, you don't know ANYTHING besides the fact that it's everywherenonzero. Then we play a game: I use that secret distribution to generate two distinct numbers, but I don't tell you either number. I then flip a fair coin to decide which of the two numbers to show you. Upon showing you the number, you're allowed to say "larger" or "smaller." If in fact the number I showed you was larger than the other number (that I didn't show you), you win if you said "larger" and lose if you said "smaller." Similarly, if the number I showed you was smaller than the other number, you win if you said "smaller" and lose if you said "larger." Obviously, you can't do this with 100% success. That's not the question. The question is whether you can come up with a strategy that's better than merely guessing at random. For example, one strategy might be to always answer that the number is the larger number. But you'd only be right 50% of the time. Is there a strategy that provides a better than 50% success rate? 


Yes. 


Maybe I should have been more clear: if so, what is one such strategy? If not, why not? 


Create a random number along the distribution. That will be your comparison value. When you turn over one of the random numbers, you will compare it to your random number. If it is higher, then you will pick "lower" and vice versa. This only gives you a slightly better than 5050 chance of being right. The reason this works is because in the off chance that you are number lies between the two numbers the other person chose, you will always be right. Otherwise, it is still just a 5050 chance. 


We used to operate a scam on the bookies some time ago. One would place a complicated bet on a large number of football teams and under stake the number of combinations. The bookies would take the bet without checking the number of doubles, trebles, four timers and so on it should be. After they would just look at the winners and settle up. One would perhaps get say a grand bet on for 700. There was no guarantee of winning but it gave one the edge. 


Well done, Steven, you are a Probability Jedi Grandmaster! 


[quote]On Apr 18, 2019, Steven Keyl wrote: Create a random number along the distribution. That will be your comparison value. When you turn over one of the random numbers, you will compare it to your random number. If it is higher, then you will pick "lower" and vice versa. This only gives you a slightly better than 5050 chance of being right. The reason this works is because in the off chance that you are number lies between the two numbers the other person chose, you will always be right. Otherwise, it is still just a 5050 chance.[/quote] Is it truly a 5050 chance otherwise? 


May I pose another similar but different question which this one brings up for me? Two real numbers are chosen at random. What is the probability that a third random number lies between them? I have no idea. 


[quote]On Apr 18, 2019, landmark wrote: May I pose another similar but different question which this one brings up for me? Two real numbers are chosen at random. What is the probability that a third random number lies between them? I have no idea.[/quote] Assuming that the choices of numbers are all three governed by the same probability distribution, then the probability is ⅓, and a little thought has convinced me that that's true irrespective of the choice of specific probability distribution. 


Sounds reasonable, but my probability intuition is often wrong in difficult cases. What would your line of thinking be on this? Edit: my brother just gave me a headslappingly easy explanation! 


[quote]On Apr 18, 2019, landmark wrote: Sounds reasonable, but my probability intuition is often wrong in difficult cases. What would your line of thinking be on this? Edit: my brother just gave me a headslappingly easy explanation![/quote] It's pretty straightforward for a uniform distribution on (0, 1) (or [0, 1]). For every other probability distribution, you can map that distribution to the uniform (0, 1) distribution via its cumulative density function. So, when you choose any three numbers from any probability distribution, you are, in essence, choosing the corresponding cumulative density numbers from a uniform distribution on [0, 1]. If the probability of the latter is ⅓, then the probability of the former is ⅓ as well. 


Thanks. Okay, one more question. Let's assumed two fixed real numbers L (low) and H (high). M is picked at random. Does the P(M): L<M<H depend on L and H? Intuition says it should, but I have a feeling it might not. 


[quote]On Apr 18, 2019, landmark wrote: Thanks. Okay, one more question. Let's assumed two fixed real numbers L (low) and H (high). M is picked at random. Does the P(M): L<M<H depend on L and H? Intuition says it should, but I have a feeling it might not.[/quote] Of course it does. 


What would be an expression then for P(M) in terms of L and H? Assume our universe is the set of all real numbers, with an equally likely probability distribution for M. 


[quote]On Apr 18, 2019, landmark wrote: What would be an expression then for P(M) in terms of L and H? Assume our universe is the set of all real numbers, with an equally likely probability distribution for M.[/quote] That one's easy: P(M) = 0. Only because for a uniform distribution on all real numbers, the probability density function is zero everywhere. So the probability of being between any two finite numbers is zero; it's the width of that segment divided by the width of the real line. 


But if P(M) =0, then it doesn't depend on L or H, which was what I was contending above. Perhaps I'm not stating the problem mathematically correctly. Assume a random process capable of generating all real numbers with equal likelihood. Is the random number more likely to be between 1 and 1000 than between 1 and 2? 


[quote]On Apr 19, 2019, landmark wrote: But if P(M) =0, then it doesn't depend on L or H, which was what I was contending above.[/quote] It's zero because you specified that the distribution for M is uniform, something you hadn't specified previously. If the distribution for M is not uniform, then P(M) will depend on H and L. [quote]On Apr 19, 2019, landmark wrote: Perhaps I'm not stating the problem mathematically correctly.[/quote] Your statement of the problem was fine. You simply added a new condition: the uniform distribution of M. [quote]On Apr 19, 2019, landmark wrote: Assume a random process capable of generating all real numbers with equal likelihood. Is the random number more likely to be between 1 and 1000 than between 1 and 2?[/quote] In that case the probability of being between 1 and 1,000 is zero, and the probability of being between 1 and 2 is also zero. However, if you have a random process capable of generating all real numbers with a normal distribution having mean 0 and variance 1, then the probability of the number being between 1 and 1,000 is bigger than the probability of it being between 1 and 2. 


For everyone so inclined, I've got a mystery that needs solving... The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it. 


[quote]On Apr 19, 2019, Steven Keyl wrote: For everyone so inclined, I've got a mystery that needs solving... The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it. [/quote] And this is where we're fortunate to have Bill. I look forward to his response to this. :) Ron 


[quote]In that case the probability of being between 1 and 1,000 is zero, and the probability of being between 1 and 2 is also zero. However, if you have a random process capable of generating all real numbers with a normal distribution having mean 0 and variance 1, then the probability of the number being between 1 and 1,000 is bigger than the probability of it being between 1 and 2.[/quote] Thanks Bill. That makes it very clear for me. I guess when I say pick a number at random, I assume the model of "Hey, think of a number," assuming we can think only of real numbers, and we think of all of them in an equally likely way. But of course, there are other assumptions that can be made. I'm thinking the same argument could be made for the set of rational numbers as opposed to the reals?but not obviously for only the set of integers. 


The only real number is One. One cannot have Two horses in reality – One can one horse and another One. One cannot have half a horse but one can have horse meat. 


[youtube]44A6ZWEOqHA[/youtube] 


[quote]On Apr 19, 2019, Steven Keyl wrote: For everyone so inclined, I've got a mystery that needs solving... The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it.[/quote] The procedure is this: discard (roughly) the first 37% of the numbers you see. (Don't forget what they are, because you'll need that; you simply remove them as candidates for being the highest.) As you continue looking at numbers, stop when you find one that's higher than any previous number: that's your choice for being the highest of the group. The percentage that you discard is, as I say, roughly 37%. Exactly, it's 1/[i]e[/i], where [i]e[/i] is the base of the natural logarithms: 2.718281828.... I don't recall the success rate of this procedure, but you're correct: it's surprisingly high. In one of John Allen Paulos' books  [b][i][url=https://www.amazon.com/InnumeracyJohnAllenPaulos/dp/B0085XTM9O/ref=sr_1_4?keywords=innumeracy&qid=1555688319&s=gateway&sr=84]Innumeracy[/url][/i][/b] I believe  the author uses this idea to formulate a plan for maximizing the probability that a person will marry the best person they can. He defines a [i]heartthrob[/i] as someone who is better than everyone who has come before him or her. You start by estimating how many people you will meet in your lifetime whom you'd consider marrying. As you go through life you discard roughly the first 37% (i.e., i/[i]e[/i]) of that estimated number, then marry the first heartthrob you meet after that group. Voilà! 


It's sometimes called the Secretary Problem: https://en.wikipedia.org/wiki/Secretary_problem 


[quote]On Apr 19, 2019, landmark wrote: It's sometimes called the Secretary Problem: https://en.wikipedia.org/wiki/Secretary_problem [/quote] I'd never heard that before, but it makes sense. Thanks! 


That's it, Bill! You found it. Thanks, also, Jack for the wiki reference. I'm going to pick up Innumeracy. Looks like a great read. Thanks for that, too! Much appreciated, gents. 


[quote]On Apr 19, 2019, Steven Keyl wrote: That's it, Bill! You found it. Thanks, also, Jack for the wiki reference. I'm going to pick up Innumeracy. Looks like a great read. Thanks for that, too! Much appreciated, gents.[/quote] My pleasure. I just looked through my copy, and I was correct that that's where I found the reference on marrying. 


Jack's link led me to read more information on "optimal stopping theory" which has applications beyond mathematics into the realm of economics and finance. Might also, by extension, have practical application in politics when formulating policy. Wow... a lot to digest. Thanks again, Jack. The book has been ordered, Bill. Looking forward to it. 


[quote]On Apr 19, 2019, Steven Keyl wrote: Jack's link led me to read more information on "optimal stopping theory" which has applications beyond mathematics into the realm of economics and finance. Might also, by extension, have practical application in politics when formulating policy. Wow... a lot to digest. Thanks again, Jack. The book has been ordered, Bill. Looking forward to it.[/quote] On the topic of politics (which you mention), in his book [b][i][url=https://www.amazon.com/BeyondNumeracyRuminationsNumbersMan/dp/0394586409/ref=sr_1_1?crid=NBNYN3C82A5X&keywords=beyond+numeracy&qid=1555713048&s=gateway&sprefix=beyond+nu%2Caps%2C205&sr=81]Beyond Numeracy[/url][/i][/b], Paulos has a fascinating discussion about voting schemes. You would find it fascinating, I have no doubt. 


In an organization I belong to, I just went through observing an election of nine candidates for three positions using [url=https://en.wikipedia.org/wiki/Single_transferable_vote ]Single Transfer Voting[/url]. What an ordeal the counting of the votes was! I'll take the word of folks who I respect that there are many democratic advantages to this system over plurality takes all voting, but it certainly is far, far from intuitive if you take a look at the link. 


[quote]On Apr 19, 2019, S2000magician wrote: [quote]On Apr 19, 2019, Steven Keyl wrote: For everyone so inclined, I've got a mystery that needs solving... The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it.[/quote] The procedure is this: discard (roughly) the first 37% of the numbers you see. (Don't forget what they are, because you'll need that; you simply remove them as candidates for being the highest.) As you continue looking at numbers, stop when you find one that's higher than any previous number: that's your choice for being the highest of the group. The percentage that you discard is, as I say, roughly 37%. Exactly, it's 1/[i]e[/i], where [i]e[/i] is the base of the natural logarithms: 2.718281828.... I don't recall the success rate of this procedure, but you're correct: it's surprisingly high. In one of John Allen Paulos' books  [b][i][url=https://www.amazon.com/InnumeracyJohnAllenPaulos/dp/B0085XTM9O/ref=sr_1_4?keywords=innumeracy&qid=1555688319&s=gateway&sr=84]Innumeracy[/url][/i][/b] I believe  the author uses this idea to formulate a plan for maximizing the probability that a person will marry the best person they can. He defines a [i]heartthrob[/i] as someone who is better than everyone who has come before him or her. You start by estimating how many people you will meet in your lifetime whom you'd consider marrying. As you go through life you discard roughly the first 37% (i.e., i/[i]e[/i]) of that estimated number, then marry the first heartthrob you meet after that group. Voilà! [/quote] Thanks Bill! I also have "Innumeracy" (along with a couple other books by Paulos), and I highly recommend them! I'll need to revisit them, as they are very insightful. Ron :) 


[quote]On Apr 20, 2019, R.S. wrote: [quote]On Apr 19, 2019, S2000magician wrote: [quote]On Apr 19, 2019, Steven Keyl wrote: For everyone so inclined, I've got a mystery that needs solving... The late Lew Brooks once told me about a method for determining the highest number of a group of numbers, the values of which are unknown at the outset. In other words, someone writes down, let's say, 10 numbers on 10 slips of paper. These can be any real numbers, and they do not need to be consecutive: 1, 10.5, 1,000,000, etc. The papers are mixed up and you start pulling out pieces of paper and reading the numbers on them. Somewhere between the 3rd and 7th piece you are able to stop and state which is the highest number in the series. If I recall, the success rate for this was ~6070%, but I may not be remembering this correctly. Does anyone know what I'm talking about? I've never heard about this anywhere else, but it may be a common problem/paradox in mathematics of which I'm unaware. I also don't remember the solution, but it was simple and straightforward. If anyone can point me in the proper direction, I'd appreciate it.[/quote] The procedure is this: discard (roughly) the first 37% of the numbers you see. (Don't forget what they are, because you'll need that; you simply remove them as candidates for being the highest.) As you continue looking at numbers, stop when you find one that's higher than any previous number: that's your choice for being the highest of the group. The percentage that you discard is, as I say, roughly 37%. Exactly, it's 1/[i]e[/i], where [i]e[/i] is the base of the natural logarithms: 2.718281828.... I don't recall the success rate of this procedure, but you're correct: it's surprisingly high. In one of John Allen Paulos' books  [b][i][url=https://www.amazon.com/InnumeracyJohnAllenPaulos/dp/B0085XTM9O/ref=sr_1_4?keywords=innumeracy&qid=1555688319&s=gateway&sr=84]Innumeracy[/url][/i][/b] I believe  the author uses this idea to formulate a plan for maximizing the probability that a person will marry the best person they can. He defines a [i]heartthrob[/i] as someone who is better than everyone who has come before him or her. You start by estimating how many people you will meet in your lifetime whom you'd consider marrying. As you go through life you discard roughly the first 37% (i.e., i/[i]e[/i]) of that estimated number, then marry the first heartthrob you meet after that group. Voilà![/quote] Thanks Bill! I also have "Innumeracy" (along with a couple other books by Paulos), and I highly recommend them! I'll need to revisit them, as they are very insightful. Ron :) [/quote] My pleasure, Ron. 


[quote]On Apr 8, 2019, landmark wrote: You asked for itit's turtles all the way down. This has been discussed before by some really amazing folks here including Tomas B and Lobowolf XXX. There are many subthreads referred to as well. Screw your head on tight because it's about to be spinning: https://www.themagiccafe.com/forums/viewtopic.php?topic=321702&forum=101 [/quote] Recently appended. 


I love this!!!!! :cheering: 